Consider a plant that cools air at (1)80% relative humitity, 30°C to (2) saturated air at 14°C. Condensate is removed at point 3, also at 14°C. If said plant receives 10m^3/min of moist air determine rates of heat and moisture removal. (See Cengel and Boles.)
Plot points 1 and 2 using the drop down box - plot a line of constant specific humidity through point 1. For each point, extract parameters from the table.
\begin{align}
\omega_1 &= 0.02158 \qquad H^*_1= 85.31kJ/kg \qquad v^*_1 = 0.8889 kg/m^3 \qquad at \; 1 \\
\omega_2 &= 0.00998 \qquad H^*_1=39.27 kJ/kg
\end{align}
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The mass flow rate of dry air is ...
$$ \dot{m}_a = \dot{V}_1/v^*_1 = 10/0.8889 = 11.25 kg/min $$The condensation rate is
$$ \dot{m}_3=\dot{m}_a (\omega_1 - \omega_2 ) = 11.25 \times (0.02158-0.00998)=0.1305kg/min$$The multistream SFEE gives the rate of heat transfer,
\begin{align} \dot{Q} &= \dot{m}_a(H^*_2-H^*_1) + \dot{m}_3 c_{pw} t_3 \\ \dot{Q} &= 11.25 \times (39.27 - 85.31) + 0.1305 \times 4.18 \times 14 kJ/min \\ \dot {Q} &= -517.95 +7.64 = -510.3 kJ/min \end{align}The all numbers agree to three significant figures with Cengel and Boles.